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3 years ago

As the frequency of sound waves increases, the wavelength of the sound waves decreases. true or false.

Physics

1 answer:

Maru [420]3 years ago

5 0

It is true that as the frequency of sound waves increases, the wavelength of the sound waves decreases.

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What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

maks197457 [2]

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

$Wavelength = \frac {h}{mv}$

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

$Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}$

$Wavelength = \frac {6.6262 * 10^{-34}}{4.859}$

Wavelength = 1.36 * 10^{-34} meters

7 0

2 years ago

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

Llana [10]

Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

a = nb

The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

$h = t \sqrt{1 - \frac{v^2}{c^2} }$

Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

So t = a = nb

and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)

i.e h = b

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

$[\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}$

$v^2 =c^2 [1- \frac{1}{n^2} ]$

$v^2 =c^2 [\frac{n^2 -1}{n^2} ]$

$v = c* \sqrt{1 - \frac{1}{n^2} }$

6 0

3 years ago

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

zaharov [31]

9,100 is the answer if you do the work so i failed for yall you welcome

8 0

3 years ago

Read 2 more answers

The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i

balandron [24]

Answer:

The answer is

Explanation:

Using Boyle's law to find the final pressure

That's

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

$P_2 = \frac{P_1V_1}{V_2}$

From the question

P1 = 115 kPa

V1 = 480 mL

V2 = 650 ml

So we have

$P_2 = \frac{115000 \times 480}{650} = \frac{55200000}{650} \\ = 84923.076923...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Which description of energy changes in nuclear reactions is correct?

Kryger [21]

The answer for this question is letter "B.Fission releases energy, and its products have greater stability."

Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition.

4 0

3 years ago